I'm 16 years old, and I'm studying for my exam maths coming this monday. In the chapter "sequences and series", there is this exercise:

Prove that a positive integer formed by $k$ times digit 1, followed by $(k-1)$
times digit 5 and ending on one 6, is the square of an integer.

I'm not a native English speaker, so my translation of the exercise might be a bit crappy. What is says is that 16, 1156, 111556, 11115556, 1111155556, etc are all squares of integers. I'm supposed to prove that. I think my main problem is that I don't see the link between these numbers and sequences.
Of course, we assume we use a decimal numeral system (= base 10)

Can anyone point me in the right direction (or simply prove it, if it is difficult to give a hint without giving the whole evidence). I think it can't be that difficult, since I'm supposed to solve it.

For sure, by using the word "integer", I mean "natural number" ($\in\mathbb{N}$)

But, I still don't see any progress in my evidence. A human being can see in these numbers a system and can tell it will be correct for $k$ going to $\infty$. But this isn't enough for a mathematical evidence.

Hint - try writing the general term in simple terms using the fact that a block of digits all the same can be summed as a geometric progression. So a sequence of $k$ '1's is $\frac{10^k-1}9 $, then see what you have.
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Mark BennetJun 10 '12 at 13:11

By the formula for the sum of a finite geometric series, we have: $$u_k=1+5 \frac{10^k-1}{9}+\frac{10^{2k}-1}{9} - \frac{10^k-1}{9}=1+ \frac{4(10^k)-4+10^{2k}-1}{9}$$
Bringing the 1 into the fraction and cancelling, we get$$u_k=\frac{10^{2k}+4(10^k)+4}{9}=\left(\frac{10^k+2}{3}\right)^2$$

Hint - try writing the general term in simple terms using the fact that a block of digits all the same can be summed as a geometric progression. So a sequence of $k$ '1's is $10^k−1\over9$, then see what you have.

Here's another way, using induction. I'm not a huge fan of proofs by induction, as they often seem to me to mask what's going on. In this case, though, induction allows you to stick to extremely elementary techniques, provided you can get your head around some notation and keep your columns in order.

It's now a matter of adding the digits in each column. A picture makes this easy:

In words: the last $k+1$ digits are just $5_{(k)}6$, as the left and middle terms have $0$'s here. In the $k+2$th position, you have $4+1=5$, from the middle and last terms; so, cumulatively, we've now got $5_{(k+1)}6$. For the next $k$ positions, the first and middle term again supply only zeroes. So, we look to the final term, which gives use $k$ $1$'s. We're now up to $1_{(k)}5_{(k+1)}6$. At this point, the final term has run out and we look to the first and middle terms to fill the $2k+2$th column. These give $9+2=11$. So, we have $111_{(k)}5_{(k+1)}6$. That is:

This is quite a nice result, and worth remembering; proving that each number in the sequence is the square of a natural number by 'guessing' what number each is the square of, and proving that this is indeed true for every number in the sequence.
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Bill MichellJun 10 '12 at 20:53